Expressions Questions and Answers

Q: Which of the following is the better approach to do the operation i = i * 16?

Correct answer: Shift left by 4 bitsShifting left by \(n\) places multiplies by \(2^n\), and \(16 = 2^4\), so i << 4 is the same as i * 16 for non-negative values - done in a single instruction rather than a multiply. In practice any modern compiler makes this substitution for you, so writing i * 16 is clearer and costs nothing.

Q: extern int s; int t; static int u; main() { } Which of s, t and u are available to a function present in another file?

Correct answer: s and tThe question is one of linkage: which of these names can be referred to from a different translation unit.extern int s; declares s with external linkage without defining it. The name refers to a single object shared across the whole program, defined in some other file, so a function in another file names the same s. Available.int t; at file scope also has external linkage. It is a tentative definition: if nothing else in the file defines t, the compiler emits a definition initialised to zero. Another file can reach it by declaring extern int t;. Available.static int u; at file scope is the exception. The static keyword here has nothing to do with lifetime - all three of these objects live for the whole run of the program - it sets internal linkage, confining the name to this translation unit. No other file can reach u, and another file may define its own unrelated u without a clash. Not available.So s and t are visible outside the file and u is not. This record was keyed to “Only s”, which overlooks that a bare file-scope definition is external by default; the option reading “S & u” has been corrected to “s and t” and the key moved to it.

Q: What is the value of the following expression? i = 1; i << 1

Correct answer: 2<< shifts the bits left, and one place left doubles the value:\[ 1 \ll 1 = 1 \times 2^1 = 2 \]The value of the expression is therefore 2. Shifting cannot produce a negative result here, since the sign bit of a small positive int is nowhere near the top.

Q: Which of the following is not an arithmetic operation?

Correct answer: a != 50;!= is a relational operator: it compares and yields 1 or 0, and does not modify a at all. The other three are compound assignment operators - a *= 20 means a = a * 20, and likewise for /= and %=.

Q: Which of the following statement is correct about the code snippet given below? #include <stdio.h> int main() { int n = 12, k; printf("%d", (k = sizeof( n + 12.0))++); return 0; }

Correct answer: The code will result in a compile time error(k = sizeof(n + 12.0))++ tries to increment the result of an assignment expression. That result is a value, not an object, so it is not a modifiable lvalue and the compiler rejects it. (Had the increment been dropped, sizeof(n + 12.0) would be 8 on a typical machine, since the int is promoted to double.)

Q: What is the output of this C code? int main() { int i = 10; int *p = &i; printf("%d\n", *p++); }

Correct answer: 10++ binds tighter than *, but it is a post-increment, so *p++ means “dereference the current p, then advance p”. The value printed is therefore i itself, 10; the pointer moves on afterwards, to an address that is no longer valid.

Q: What is the output of this C code? void main() { int a = 5; int b = ++a + a++ + --a; printf("Value of b is %d", b); }

Correct answer: Undefined behaviourb = ++a + a++ + --a; modifies a three times within one expression, with no sequence point between the changes. The standard leaves the result undefined - not merely unspecified - so any answer a particular compiler gives is accidental. This is why such expressions must be split into separate statements.

Q: What is the output of the following program? #include <stdio.h> int main() { static int a = 3; printf("%d", a--); return 0; }

Correct answer: 3a-- is a post-decrement: the current value is what the expression yields, and the decrement happens afterwards. So printf receives 3; a becomes 2 immediately after, but nothing prints it. The static qualifier matters only if the function is called again.

Q: Longevity of a variable refers to

Correct answer: The duration for which the variable retains a given value during the execution of a program.Longevity (or lifetime, or storage duration) is about time - how long the variable’s storage exists and keeps its value. That is a different question from scope, which is about place: the part of the program in which the name is visible. A static local has a short scope but the longest possible lifetime.

Q: Which of the following statement are correct? (i) The maximum value a variable can hold depends upon its storage class. (ii) By default all variables enjoy a static storage class.

Correct answer: Both I & II are incorrect(i) is false: the range of a variable depends on its type - char, int, long - not on where it is stored. A static int and an auto int hold exactly the same range. (ii) is false: the default for a variable declared inside a function is auto, not static; only file-scope variables have static storage duration.

21.

Which of the following is the better approach to do the operation i = i * 16?

A.Multiply i by 16 and keep it
B.Shift left by 4 bits
C.Add i to itself 16 times
D.Shift right by 4 bits

Answer: B

Correct answer: Shift left by 4 bits

Shifting left by \(n\) places multiplies by \(2^n\), and \(16 = 2^4\), so i << 4 is the same as i * 16 for non-negative values - done in a single instruction rather than a multiply. In practice any modern compiler makes this substitution for you, so writing i * 16 is clearer and costs nothing.

Discuss in Forum

22.

extern int s;
int t;
static int u;

main()
{

}

Which of s, t and u are available to a function present in another file?

A.Only s
B.s and t
C.S, t, u
D.None

Answer: A

Correct answer: s and t

The question is one of linkage: which of these names can be referred to from a different translation unit.

extern int s; declares s with external linkage without defining it. The name refers to a single object shared across the whole program, defined in some other file, so a function in another file names the same s. Available.

int t; at file scope also has external linkage. It is a tentative definition: if nothing else in the file defines t, the compiler emits a definition initialised to zero. Another file can reach it by declaring extern int t;. Available.

static int u; at file scope is the exception. The static keyword here has nothing to do with lifetime - all three of these objects live for the whole run of the program - it sets internal linkage, confining the name to this translation unit. No other file can reach u, and another file may define its own unrelated u without a clash. Not available.

So s and t are visible outside the file and u is not. This record was keyed to “Only s”, which overlooks that a bare file-scope definition is external by default; the option reading “S & u” has been corrected to “s and t” and the key moved to it.

Discuss in Forum

23.

What is the value of the following expression?

i = 1;
i << 1 < 1>

A.2
B.-2
C.1
D.-1

Answer: A

Correct answer: 2

<< shifts the bits left, and one place left doubles the value:

\[ 1 \ll 1 = 1 \times 2^1 = 2 \]

The value of the expression is therefore 2. Shifting cannot produce a negative result here, since the sign bit of a small positive int is nowhere near the top.

Discuss in Forum

24.

Which of the following is not an arithmetic operation?

A.a *= 20;
B.a /= 30;
C.a %= 40;
D.a != 50;

Answer: D

Correct answer: a != 50;

!= is a relational operator: it compares and yields 1 or 0, and does not modify a at all. The other three are compound assignment operators - a *= 20 means a = a * 20, and likewise for /= and %=.

Discuss in Forum

25.

Which of the following statement is correct about the code snippet given below?

#include <stdio.h>
int main()
{
int n = 12, k;
printf("%d", (k = sizeof( n + 12.0))++);
return 0;
}

A.The code will print 17
B.The code will print 5
C.The code will result in a compile time error
D.The code will print 4

Answer: C

Correct answer: The code will result in a compile time error

(k = sizeof(n + 12.0))++ tries to increment the result of an assignment expression. That result is a value, not an object, so it is not a modifiable lvalue and the compiler rejects it. (Had the increment been dropped, sizeof(n + 12.0) would be 8 on a typical machine, since the int is promoted to double.)

Discuss in Forum

26.

What is the output of this C code?

int main()
    {
        int i = 10;
        int *p = &i;
        printf("%d\n", *p++);
    }

A.10
B.11
C.Garbage value
D.Address of i

Answer: A

Correct answer: 10

++ binds tighter than *, but it is a post-increment, so *p++ means “dereference the current p, then advance p”. The value printed is therefore i itself, 10; the pointer moves on afterwards, to an address that is no longer valid.

Discuss in Forum

27.

What is the output of this C code?

void main()
    {
        int a = 5;
        int b = ++a + a++ + --a;
        printf("Value of b is %d", b);
    }

A.Value of b is 16
B.Value of b is 21
C.Value of b is 15
D.Undefined behaviour

Answer: D

Correct answer: Undefined behaviour

b = ++a + a++ + --a; modifies a three times within one expression, with no sequence point between the changes. The standard leaves the result undefined - not merely unspecified - so any answer a particular compiler gives is accidental. This is why such expressions must be split into separate statements.

Discuss in Forum

28.

What is the output of the following program?

#include <stdio.h>

int main()
{
static int a = 3;
printf("%d", a--);
return 0;
}

Answer: C

Correct answer: 3

a-- is a post-decrement: the current value is what the expression yields, and the decrement happens afterwards. So printf receives 3; a becomes 2 immediately after, but nothing prints it. The static qualifier matters only if the function is called again.

Discuss in Forum

29.

Longevity of a variable refers to

A.The duration for which the variable retains a given value during the execution of a program.
B.The portion of a program in which the variable may be visible.
C.Internal linkage of a variable.
D.External linkage of a variable.

Answer: A

Correct answer: The duration for which the variable retains a given value during the execution of a program.

Longevity (or lifetime, or storage duration) is about time - how long the variable’s storage exists and keeps its value. That is a different question from scope, which is about place: the part of the program in which the name is visible. A static local has a short scope but the longest possible lifetime.

Discuss in Forum

30.

Which of the following statement are correct?

(i) The maximum value a variable can hold depends upon its storage class.
(ii) By default all variables enjoy a static storage class.

A.Only I is correct
B.Only II is correct
C.Both I & II are correct
D.Both I & II are incorrect

Answer: D

Correct answer: Both I & II are incorrect

(i) is false: the range of a variable depends on its type - char, int, long - not on where it is stored. A static int and an auto int hold exactly the same range. (ii) is false: the default for a variable declared inside a function is auto, not static; only file-scope variables have static storage duration.

Discuss in Forum

Loading…

Topics

Other Subjects

Tags: Expressions Questions and Answers || Expressions MCQ Questions and Answers || Expressions GK Questions and Answers || Expressions GK MCQ Questions || Expressions Multiple Choice Questions and Answers || Expressions GK || GK on Expressions || C Programming Questions and Answers || C Programming MCQ Questions and Answers || C Programming GK Questions and Answers || GK on C Programming

Expressions Questions and Answers

Topics

Other Subjects

About Us

Stay informed about all important Dates & Notifications